Two identical springs are connected in series and parallel as shown in the figure. If fs and fp are frequencies of arrangements, what is (fs/fp) ?

Question

Two identical springs are connected in series and parallel as shown in the figure. If fs and fp are frequencies of arrangements, what is (fs/fp) ? <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-dy23ttcpaih8talp.jpg" /> (A) 1:2 (B) 2:1 (C) 1:3 (D) 3:1

Tardigrade Admin · Accepted Answer

In first case, springs are connected in parallel, so their equivalent spring constant

k_{p} = k_{1} + k_{2}

So, frequency of this spring-block system is

f_{p} = \frac{1}{2 π} \frac{k _{p}}{m}

or

f_{p} = \frac{1}{2 π} \frac{k _{1} + k _{2}}{m}

but

k_{1} = k_{2} = k

∴

f_{p} = \frac{1}{2 π} \frac{2 k}{m}

..(i)
Now in second case, springs are connected in series, so their equivalent spring constant

k = \frac{k _{1} k _{2}}{k _{1} + k _{2}}

Hence, frequency of this arrangement is given by

f_{s} = \frac{1}{2 π} \frac{k _{1} k _{2}}{( k _{1} + k _{2} ) m}

or

f_{s} = \frac{1}{2 π} \frac{k}{2 m}

...(ii)
Dividing Eq. (ii) by Eq. (i), we get

\frac{f _{s}}{f _{p}} = \frac{\frac{1}{2 π} \frac{k}{2 m}}{\frac{1}{2 π} \frac{2 k}{m}} = \frac{1}{4}

or

\frac{f _{s}}{f _{p}} = \frac{1}{2}

Q. Two identical springs are connected in series and parallel as shown in the figure. If fs​andfp​ are frequencies of arrangements, what is fp​fs​​ ?

1:2

2:1

1:3

3:1

Q. Two identical springs are connected in series and parallel as shown in the figure. If $f_{s} an d f_{p}$ are frequencies of arrangements, what is $\frac{f _{s}}{f _{p}}$ ?